Re: Per line ${..//..}

[DervishD <zsh@xxxxxxxxxxxx>]

    Hi Peter :)

 * Peter Stephenson <pws@xxxxxxx> dixit:
> DervishD wrote:
> >     I have a parameter containing a bunch of lines, and I have to get
> > rid of comments (that is "$line" == (#s)[" $'\t'"]#\#*(#e)) and empty
> > lines (that is "$line" == (#s)[" $'\t'"]#(#e)). I've tried using the
> > ${..//..} syntax for doing substitutions in parameters, but with no
> > success because I don't know how to properly mark the beginning and
> > end of a line.
> 
> You can do something like:
> 
> % lines=(foo "# comment" bar "    ")
> % print -l $lines
> foo
> # comment
> bar
> 
> % print -l ${lines:#[[:space:]]#(\#*|)}
> foo
> bar

    Unfortunately the original data is not in an array :( Will this
work?:

    olddata=(${(f)olddata))
    olddata=(${olddata:#...})

    I completely forgot about the ${NAME:#PATTERN}. Better said, I
was stuck with ${..//..}. Thanks a lot for the suggestion. You've
given me another idea... I can convert all parameters that store file
contents into arrays, although some of them have \C-M as newline
markers, but that's easy to solve. This way I can do per-line
processing without the need for loops.

    Raúl Núñez de Arenas Coronado

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